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Pertanyaan

Perhatikan gambar di bawah ini pernyataan yang benar adalah...

  1. koefisien x squared adalah -1

  2. koefisien x adalah -2

  3. Titik pot dengan sb y space equals left parenthesis 0 comma 4 right parenthesis

  4. Nilai min 4

Pembahasan Soal:

Bentuk umum fungsi kuadrat adalah straight f left parenthesis straight x right parenthesis space equals space ax ² space plus space bx space plus space straight c space dengan space straight a space not equal to space 0. space space

Jika a space greater than space 0 maka kurva terbuka ke atas dan jika a < 0 maka kurva terbuka ke bawah

Ada beberapa cara untuk menentukan persamaan fungsi kuadrat, yaitu

Jika diketahui titik puncak (xp, yp) dan melalui titik (x, y)

y space equals space a left parenthesis x space – space x p right parenthesis ² space plus space y p

Jika diketahui titik potong terhadap sumbu x di titik (x₁, 0) dan (x₂, 0) serta melalui titik (x, y)

y space equals space a left parenthesis x space – space x ₁ right parenthesis left parenthesis x space – space x ₂ right parenthesis

Jka dilihat dari grafiknya terbuka ke bawah mana nilai a < 0.

Kurva memotong sumbu x di titik (–1, 0) dan (3, 0).

a. sumbu simetri

table attributes columnalign right center left columnspacing 0px end attributes row cell straight x subscript straight p space end cell equals cell space 1 half space left parenthesis straight x ₁ space plus space straight x ₂ right parenthesis space space end cell row cell straight x subscript straight p space end cell equals cell 1 half space left parenthesis – 1 space plus space 3 right parenthesis space end cell row cell space straight x subscript straight p space end cell equals cell space 1 half space left parenthesis 2 right parenthesis space space end cell row cell straight x subscript straight p space end cell equals cell space 1 end cell end table

b. rumus fungsi kuadrat

 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight y space end cell equals cell space straight a left parenthesis straight x space – space straight x ₁ right parenthesis left parenthesis straight x space – space straight x ₂ right parenthesis space end cell row cell space straight y space end cell equals cell space straight a left parenthesis straight x space – space left parenthesis – 1 right parenthesis right parenthesis left parenthesis straight x space – space 3 right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis straight x space plus space 1 right parenthesis left parenthesis straight x space – space 3 right parenthesis space end cell row cell space straight y space end cell equals cell space straight a left parenthesis straight x ² space – space 3 straight x space plus space straight x space – space 3 right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis straight x ² space – space 2 straight x space – space 3 right parenthesis end cell end table

c. nilai optimum

Diperoleh dengan mensubstitusikan

table attributes columnalign right center left columnspacing 0px end attributes row cell straight y space end cell equals cell space straight a left parenthesis straight x space – space straight x ₁ right parenthesis left parenthesis straight x space – space straight x ₂ right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis straight x space – space left parenthesis – 1 right parenthesis right parenthesis left parenthesis straight x space – space 3 right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis straight x space plus space 1 right parenthesis left parenthesis straight x space – space 3 right parenthesis space end cell row cell space straight y space end cell equals cell space straight a left parenthesis straight x ² space – space 3 straight x space plus space straight x space – space 3 right parenthesis space end cell row cell space straight y space end cell equals cell space straight a left parenthesis straight x ² space – space 2 straight x space – space 3 right parenthesis end cell row cell straight y subscript straight p space space end cell equals cell space straight a left parenthesis xp ² space – space 2 xp space – space 3 right parenthesis space space end cell row cell straight y subscript straight p space space end cell equals cell space straight a left parenthesis 1 ² space – space 2 left parenthesis 1 right parenthesis space – space 3 right parenthesis space space end cell row cell straight y subscript straight p space space end cell equals cell space straight a left parenthesis 1 space – space 2 space – space 3 right parenthesis space end cell row cell space straight y subscript straight p space space end cell equals cell space straight a left parenthesis – 4 right parenthesis space space end cell row cell straight y subscript straight p space end cell equals cell space – 4 straight a space end cell end table

d. titik puncak

left parenthesis x subscript p comma space y subscript p right parenthesis space space equals space left parenthesis 1 comma space – 4 a right parenthesis

e. titik potong grafik dengan sumbu y  

Diperoleh jika x = 0

 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight y space end cell equals cell space straight a left parenthesis straight x ² space – space 2 straight x space – space 3 right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis 0 ² space – space 2 left parenthesis 0 right parenthesis space – space 3 right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis 0 space – space 0 space – space 3 right parenthesis space space end cell row cell straight y space end cell equals cell space straight a left parenthesis – 3 right parenthesis space space end cell row cell straight y space end cell equals cell space – 3 straight a space space left parenthesis 0 comma space – 3 straight a right parenthesis end cell end table

Jadi agar diperoleh nilai a, dalam soal harus diketahui titik lainnya.  

Kita misalkan nilainya adalah a = –1

maka

b. rumus fungsi kuadrat

table attributes columnalign right center left columnspacing 0px end attributes row cell straight y space end cell equals cell space straight a left parenthesis straight x ² space – space 2 straight x space – space 3 right parenthesis space end cell row blank equals cell space – 1 left parenthesis straight x ² space – space 2 straight x space – space 3 right parenthesis space end cell row blank equals cell space – straight x ² space plus space 2 straight x space plus space 3 end cell end table

jadi, fungsi kuadratnya adalah y space equals space minus x squared plus 2 x plus 3.

c. nilai optimum

y subscript p space equals space – 4 a space equals space – 4 left parenthesis – 1 right parenthesis space equals space 4

Nilai optimumnya adalah 4

d. titik puncak

left parenthesis x subscript p comma space y subscript p right parenthesis space equals space left parenthesis 1 comma space 4 right parenthesis

e. titik potong grafik dengan sumbu y  

left parenthesis 0 comma – 3 a right parenthesis space equals space left parenthesis 0 comma space – 3 left parenthesis – 1 right parenthesis right parenthesis space equals space left parenthesis 0 comma space 3 right parenthesis

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

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